Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.4 - The Chain Rule - Exercises 14.4 - Page 816: 7

Answer

dz/du = dz/dx * dx/du = 4e^x cos y * (1/u) = 4 cos y / u dz/dy = dz/dx * dx/dy + dz/dy = 4e^x cos y * -u sin y + 4 sin y = 4(cos y - u sin y)

Work Step by Step

a) Expressing z directly in terms of u and y: z = 4ex ln y = 4e^(ln(u cos y)) ln y = 4u cos y ln y By using the Chain Rule: dz/du = dz/dx * dx/du = 4e^x cos y * (1/u) = 4 cos y / u dz/dy = dz/dx * dx/dy + dz/dy = 4e^x cos y * -u sin y + 4 sin y = 4(cos y - u sin y) 0z>0u = 4 cos y / u 0z>0y = 4(cos y - u sin y) b) Evaluating 0z>0u and 0z>0y at the given point (u, y): u = 2, y = p>4, x = ln (2 cos y) 0z>0u = 4 cos y / u = 4 cos (p>4) / 2 0z>0y = 4(cos y - u sin y) = 4(cos (p>4) - 2 sin (p>4))
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